How do you differentiate  y= sqrt((3x)/(2x-3)) using the chain rule?

Apr 6, 2018

The final answer, using the chain rule and the quotient rule is:

f'(x)=9*(2x-3)^(5/2)/(2*(3x)^(1/2).

See below for the details.

Explanation:

I find this type of function much easier to handle without the radical sign, so I will rewrite it as $f \left(x\right) = {\left(\frac{3 x}{2 x - 3}\right)}^{\frac{1}{2}}$.

As usual with the chain rule, we begin with the "outermost" function (i.e., ask yourself, "If I wanted to find f(2), for example, what is the last function that I would do?). In this case, that is clearly the rational exponent (formerly the radical). Thus, I differentiate that first using the power rule, then handle the rational function using the quotient rule.

So: the answer becomes:

$f ' \left(x\right) = \left(\frac{1}{2}\right) \cdot {\left(\frac{3 x}{2 x - 3}\right)}^{- \frac{1}{2}} \cdot \frac{\left(2 x - 3\right) \cdot 3 - \left(3 x \cdot 2\right)}{2 x - 3} ^ 2$

$f ' \left(x\right) = \left(\frac{1}{2}\right) \cdot {\left(\frac{3 x}{2 x - 3}\right)}^{- \frac{1}{2}} \cdot \frac{6 x - 9 - 6 x}{2 x - 3} ^ 2$

$f ' \left(x\right) = \left(\frac{1}{2}\right) \cdot {\left(3 x\right)}^{- \frac{1}{2}} / {\left(2 x - 3\right)}^{- \frac{1}{2}} \cdot \frac{9}{2 x - 3} ^ 2$

f'(x)=9*(2x-3)^(5/2)/(2*(3x)^(1/2)